Finding General Solution to ODE w/A(x) & y(x)

In summary: AB+AB''+A'B'+B/A)=0The equation becomes ##z''AB+z(AB''+A'B'+B/A)=0##. Now I have to replace the B and its derivatives in that.
  • #1
fluidistic
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Homework Statement


Find the general solution to ##A(x)y''+A'(x)y'+\frac{y}{A(x)}=0## where A(x) is a known function and y(x) is the unknown one.
Hint:Eliminate the term that contains the first derivative.


Homework Equations


Not sure.


The Attempt at a Solution


So I don't really know how to tackle it. I guess that the hint suggests a change of variable that would get rid of the y' term. So I tried z=y'A, z=Ay and z=y/A. All failed to express the ODE as a function of z and its derivative(s) and A and its derivative(s).
I am therefore stuck. Any other hint will be welcome!
 
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  • #2
What if you try the following substition [itex]y=Bz[/itex], where B is a known function that is yet to be decided.

What do you get if you perform this substitution?

What should we take for B if we want the z' term to vanish?
 
  • #3
micromass said:
What if you try the following substition [itex]y=Bz[/itex], where B is a known function that is yet to be decided.

What do you get if you perform this substitution?
Good idea! ##x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0##.

What should we take for B if we want the z' term to vanish?
##B=Ce^{-x}/A##. I checked out and indeed the terms in front of ##z'## vanishes.
I'm left to solve ##z''A'Ce^{-x} \left ( \frac{1}{A}+ \frac{A'}{A^2} \right ) + z \left ( \frac{A'}{A} + \frac{A'^2}{A^2}- \frac{Ce^{-x}}{A^2} \right ) =0##. That really does not look beautiful/easy to me. :eek:
 
  • #4
fluidistic said:
Good idea! ##x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0##.

I get something different here. I get that the term of z' is
[tex]A^\prime B + 2A B^\prime[/tex]

The form of my B is also much simpler.
 
  • #5
micromass said:
I get something different here. I get that the term of z' is
[tex]A^\prime B + 2A B^\prime[/tex]

The form of my B is also much simpler.
I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so ##B=kA^{-1/2}## where k is a constant.

Edit: I might have rushed through the math but I reach that ##z''+z \left ( 1-\frac{2}{A'} \right ) =0##. Which has different solutions depending on the sign of the term in front of z. I guess I made another mistakes.
 
Last edited:
  • #6
fluidistic said:
I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so ##B=kA^{-1/2}## where k is a constant.

OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

Anyway, with that choice of B, what does your equation become then?
 
  • #7
micromass said:
OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

Anyway, with that choice of B, what does your equation become then?
I now get ##B'=-\frac{1}{2}A^{-3/2}##, ##B''=\frac{3}{4}A^{-5/2}A'^2-\frac{A^{-3/2}A''}{2}##.
The equation becomes ##z''AB+z(AB''+A'B'+B/A)=0##. Now I have to replace the B and its derivatives in that.
Edit: I've just done it and it's not beautiful so far.
 

1. What is the general solution to an ordinary differential equation with A(x) and y(x)?

The general solution to an ordinary differential equation with A(x) and y(x) is a mathematical expression that includes all possible solutions to the equation. It takes the form of y(x) = C + f(x), where C is a constant and f(x) is a function of x. This solution can be used to find the specific solution for a given initial condition.

2. How do you find the general solution to an ODE with A(x) and y(x)?

To find the general solution to an ODE with A(x) and y(x), you can use various methods such as separation of variables, substitution, or integrating factors. These methods involve manipulating the equation and solving for y(x) in terms of x. Once you have the general solution, you can use it to find the specific solution for a given initial condition.

3. Can the general solution to an ODE with A(x) and y(x) be expressed in a closed form?

Yes, the general solution to an ODE with A(x) and y(x) can be expressed in a closed form. This means that it can be written as a finite combination of known functions, such as polynomials, trigonometric functions, or exponential functions. However, in some cases, the general solution may involve special functions that cannot be expressed in a closed form.

4. What is the role of A(x) in finding the general solution to an ODE with A(x) and y(x)?

A(x) is a function that represents the coefficient of the derivative of y(x) in the ODE. It plays a crucial role in finding the general solution because it affects the methods used to solve the equation and the form of the solution. A(x) can also provide information about the behavior of the solution, such as whether it is increasing or decreasing.

5. How do you use the general solution to an ODE with A(x) and y(x) to find a specific solution?

To find a specific solution, you need to use the initial condition given in the problem. Plug in the initial value for x and y(x) into the general solution, and then solve for the constant C. This will give you a particular solution that satisfies both the ODE and the initial condition. Alternatively, you can also use the general solution to graph the family of solutions and visually determine the specific solution that passes through the given point.

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